For the function \(f(x) = \displaystyle\frac{x}{\log x}\), \(x \gt 0\), \(x \ne 1\):
(A) \(f'(x) = \displaystyle\frac{\log x - 1}{(\log x)^{2}}\).
(B) f is strictly increasing in (1, e).
(C) f is strictly decreasing in (0, 1).
(D) The local minimum value of f is e.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A), (C) and (D) only
By the quotient rule:
\( \displaystyle f'(x) = \frac{\log x \cdot 1 - x \cdot \frac{1}{x}}{(\log x)^{2}} = \frac{\log x - 1}{(\log x)^{2}} \)
The denominator is positive for x ≠ 1, so the sign of f′(x) is the sign of log x − 1:
\(f'(x) \lt 0\) for \(0 \lt x \lt 1\) and for \(1 \lt x \lt e\)
\(f'(x) \gt 0\) for \(x \gt e\)
At x = e, f′ changes from − to +, so by the first derivative test there is a local minimum, with value
\( \displaystyle f(e) = \frac{e}{\log e} = e \)
So f decreases in (0, 1) and in (1, e), increases in (e, ∞), and has local minimum value e at x = e.
Checking the statements
(A) \(f'(x) = \displaystyle\frac{\log x - 1}{(\log x)^{2}}\). — Correct. Quotient rule with \((\log x)' = \frac{1}{x}\).
(B) f is strictly increasing in (1, e). — Incorrect. For 1 < x < e, log x < 1, so f′(x) < 0 and f is decreasing.
(C) f is strictly decreasing in (0, 1). — Correct. For 0 < x < 1, log x < 0, so log x − 1 < 0 and f′(x) < 0.
(D) The local minimum value of f is e. — Correct. f′ changes from − to + at x = e, and f(e) = e.